Question

A piston/cylinder arrangement with a linear spring similar to Fig. P3.105 contains R-134a at 15◦C, x= 0.4 and a volume of 0.02m3. It is heated to 60◦C, at which point the specific volume is 0.030 02 m3/kg. Find the final pressure, thework, and the heat transfer in the process.

Answer 1

Answer:

The answers the question are

Final pressure p₂ is 18.166 kPa

The work done, W is -0.192 kJ/kg

The heat transfer, Q is 48.052 kJ/kg

Explanation:

A force balance reveals that the gas is at equilibrium with the spring piston. That is when the temperature increases results in the increase in volume and the force exerted on the piston is constant. Therefore the system is a constant pressure system

T₁ = 15 °C,  V₁ = 0.02 m³,  

u₁ = $u_{f1} + x*u_{fg1}$,  From R-134a tables

u₁ = 71.93 + 0.4×238.39 = 238.39 kJ/kg

u₁ = 238.39 kJ/kg

The mass of R-134a in the piston cylinder arrangement

m = $\frac{V_1}{v_1}$

Where

V₁ = Volume of R-134a and

v₁ = Specific volume at state 1

v₁ is given by v₁ = $v_{f1} + x*v_{fg1}$ = 0.804/(10³) + (0.04211 - 0.8041/10³)

= 0.01732646 m³ = 1.733 × 10⁻² m³

Therefore m = $\frac{0.02}{0.01732646}$ = 1.1543 kg

h₁ = $h_{f1} + x*h_{fg1}$ =72.32 + 0.4×(258.97 - 72.32) = 146.98 kJ/kg

Similarly s₁ =$s_{f1} + x*s_{fg1}$ = 0.2768 + 0.4×(0.9245-0.2768)

= 0.53588 kJ/kg-K

u₂ = 137.79 +0.4×(259.25 - 137.79) = 189.374 kJ/kg

h₂ = 139.38 + 0.4×(278.51 - 139.38) = 195.032 kJ/kg

v₂ = (0.948)/10³ + 0.4 × (0.01144 - (0.948)/10³) = 0.0051448 m³

v₂ = 0.030 m³ then V₂ = m×v₂ = 3.465×10³ m³

For a constant pressure process

Q + W = u₂ - u₁ = 186.374 - 138.514 = 47.86 kJ/kg

Q = h₂ - h₁ = 195.032 - 146.98 = 48.052 kJ/kg

W = Δu - Q = 47.86 -48.052 =  -0.192 kJ/kg

From W = p·dv we have W = p×(5.145×10⁻³-1.733×10⁻²) =-1.22×10⁻²×p

That is  -0.192 kJ/kg×1.1543 kg = -1.22×10⁻²×p or p = 18.166 kPa